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By Marcel Proust, a cura di Paolo Pinto e Giuseppe Grasso

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What we now prove is an axiom corresponding to this rule. So, assume A set, B(x) set (x ∈ A), C(z) set (z ∈ (Σx ∈ A) B(x)) and let f ∈ (Πx ∈ A) (Πy ∈ B(x)) C((x, y)). We want to find an element of (Πx ∈ A) (Πy ∈ B(x)) C((x, y)) → (Πz ∈ (Σx ∈ A) B(x)) C(z). We define Ap(f, x, y) ≡ Ap(Ap(f, x), y) for convenience. Then Ap(f, x, y) is a ternary function, and Ap(f, x, y) ∈ C((x, y)) (x ∈ A, y ∈ B(x)). So, assuming z ∈ (Σx ∈ A) B(x), by Σ-elimination, we obtain E(z, (x, y) Ap(f, x, y)) ∈ C(z) (discharging x ∈ A and y ∈ B(x)), and, by λ-abstraction on z, we obtain the function (λz) E(z, (x, y) Ap(f, x, y)) ∈ (Πz ∈ (Σx ∈ A) B(x)) C(z) with argument f .

Now by abstraction on w, z, y, x in that order, we obtain a proof of the claim: (x ∈ A) (z ∈ I(A, x, y)) x=y∈A B(x) set (w ∈ B(x)) B(x) = B(y) w ∈ B(y) (λw) w ∈ B(x) ⊃ B(y) (λz) (λw) w ∈ I(A, x, y) ⊃ (B(x) ⊃ B(y)) (λx) (λy) (λz) (λw) w ∈ (∀x ∈ A) (∀y ∈ A) I(A, x, y) ⊃ (B(x) ⊃ B(y)) The same problem (of justifying Leibniz’s principle) was solved in Principia by the use of impredicative second order quantification. There one defines (a = b) ≡ (∀X) (X(a) ⊃ X(b)) from which Leibniz’s principle is obvious, since it is taken to define the meaning of identity.

Instead, we convince ourselves directly of its consistency in the following simple minded way. (2) Simple minded consistency. This means simply that ⊥ cannot be proved, or that we shall never have the right to judge ⊥ true (which unlike the proposition Cons above, is not a mathematical proposition). To convince ourselves of this, we argue as follows: if c ∈ ⊥ would hold for some element (construction) c, then c would yield a canonical element d ∈ ⊥; but this is impossible since ⊥ has no canonical element by definition (recall that we defined ⊥ ≡ N0 ).

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Alla ricerca del tempo perduto. Il Tempo ritrovato by Marcel Proust, a cura di Paolo Pinto e Giuseppe Grasso


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